關於Quadratic Equation的A.Maths.Q

a) sub α = 2 to x^3 + 6x – 20 = 0 :
2^3 + 6(2) - 20 = 0
8+12-20 = 0 is true.
Since It is known that α is the only real root , so α = 2.
b) 1 /α + 1 /β + 1 /γ
= 1/2 + 1 /β + 1 /γ
= 1/2 + ( β + γ) / (β γ)
x^3 + 6x – 20 = (x-2)(x^2 + 2x + 10)
βandγare the roots of x^2 + 2x + 10 = 0
β + γ = - 2
βγ = 10
So 1/2 + ( β + γ) / (β γ)
= 1/2 + (-2/10)
= 3/10

(a) Let f(x) = x^3 + 6x - 20
f(2) = 2^3 + 6 (2) - 20 = 8 + 12 - 20 = 0
Therefore x = 2 is a solution of x^3 + 6x – 20 = 0

(b)
α + β + γ = 0
αβ + βγ + γα = 6
αβγ = 20

1 / α + 1 / β + 1 / γ
= ( βγ + γα + αβ ) / ( αβγ )
= 6 / 20
= 3 / 10

1. Given thatα，β andγare the roots of the equation x^3 + 6x – 20 = 0
It is known that α is the only real root.
(a) Prove that α= 2.
x^3+6x－20=0
x^3－2x^2+2x^2+6x－20=0
x^2(x－2)+(x－2)(2x+10)=0
(x^2+2x+10)(x－2)=0
x^2+2x+10=(x+1)^2+9>0 無實根
So x=2 即α = 2

2009-10-09 23:46:48 補充：
(b) Evaluate 1/α+ 1/β+ 1/γ.
設y=1/x =>x=1/y
x^3 + 6x – 20 = 0.
(1/y)^3+6(1/y)－20=0
1+6y^2－20y^3=0
20y^3－6y^2－1=0
So 1/α，1/β，1，/γ為20y^3－6y^2－1=0之三根
1/α+1/β+1 /γ=－(－6)/20=3/10